I'm having a little bit of confusion with the following. The integral is given as:

$$\int_0^1 x^{-1/2} dx = [2x^{1/2}]_0^1 = 2$$

But from what I understand, a Riemann integrable function must be bounded. How does one reconcile this with the fact that

$$x^{-1/2} \to \infty \text{ as }x \to 0$$

I suppose a natural follow up question would be: If the integral of a function is finite, is it Riemann integrable?

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    $\begingroup$ There's a fuller explanation here (math.stackexchange.com/questions/610054/…. The short version is that Riemann integrable does imply bounded, but in the case of unbounded integrands like this one, you can regard $\int_0^1 x^{-1/2}\, dx$ as shorthand for $\lim_{\epsilon \to 0} \int_\epsilon^1 x^{-1/2}\, dx$, which presents no difficulty. $\endgroup$ – Connor Harris Jan 2 '19 at 2:23
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    $\begingroup$ I'm a bit confused by your followup question of "If the integral of a function is finite, is it Riemann integrable". The first part basically assumes an "integral" exists (I assume Riemann rather than Lebesgue or some other type), at least being "finite", to ask in the second part if it exists. You may wish to rephrase this to make it more clear. $\endgroup$ – John Omielan Jan 2 '19 at 2:29
  • $\begingroup$ @ConnorHarris So does this mean that it is Improper Riemann integrable in this case? $\endgroup$ – Sean Lee Jan 2 '19 at 2:30
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    $\begingroup$ Yes, it would be improper Riemann integrable. $\endgroup$ – Connor Harris Jan 2 '19 at 2:31
  • $\begingroup$ @JohnOmielan I was referring to I guess what would be considered the 'high school' method of calculating integrals. $\endgroup$ – Sean Lee Jan 2 '19 at 2:32

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